Question

An SAT prep course claims to improve the test score of students. The table below shows the scores for seven students the first two times they took the verbal SAT. Before taking the SAT for the second time, each student took a course to try to improve his or her verbal SAT scores. Do these results support the claim that the SAT prep course improves the students' verbal SAT scores?

Let d=(verbal SAT scores prior to taking the prep course)−(verbal SAT scores after taking the prep course). Use a significance level of α=0.1 for the test. Assume that the verbal SAT scores are normally distributed for the population of students both before and after taking the SAT prep course.

Student	1	2	3	4	5	6	7
Score on first SAT	510	490	540	410	520	550	510
Score on second SAT	530	540	560	490	550	590	530

Step 1 of 5:

State the null and alternative hypotheses for the test

Step 2 of 5:

Find the value of the standard deviation of the paired differences. Round your answer to one decimal place.

Step 3 of 5:

Compute the value of the test statistic. Round your answer to three decimal places.

Step 4 of 5:

Determine the decision rule for rejecting the null hypothesis H0H0. Round the numerical portion of your answer to three decimal places.

Step 5 of 5:

Make the decision for the hypothesis test.

Answer 1

Number	1 SAT	2 SAT	Difference	( di - d̅ )2
	510	530	-20	293.8776
	490	540	-50	165.3061
	540	560	-20	293.8776
	410	490	-80	1836.7347
	520	550	-30	51.0204
	550	590	-40	8.1633
	510	530	-20	293.8776
Total	3530	3790	-260	2942.8571

Step1

To Test :-

H0 :- µd = 0

H1 :- µd < 0

Step 2

S(d) = √(Σ (di - d̅)2 / n-1)
S(d) = √(2942.8571 / 7-1) = 22.1

Step 3

t = d̅ / ( S(d) / √(n) )
t = -37.1429 / ( 22.1467 / √(7) )
t = -4.437

Step 4

Reject null hypothesis if t < - t(α)
Critical value t(α) = t(0.1) = 1.440
t < 1.440

Step 5

t < - t(α) = -4.4373 < -1.4398
Result :- Reject null hypothesis

There is sufficient evidence to support the claim that student took a course to improve his or her verbal SAT scores.